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I want to find a formula for the coefficients of the $x^{n-2}$ and $x^{n-1}$ terms of the chebyshev polynomial $T_0(x) = 1, T_1(x) = x; T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x)$. We've already shown the leading coefficient in $T_n$ is $2^{n-1}$, but i'm not sure how we find the formula for what I want.

Can anyone help me with this?

Taln
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    Are you sure about the recursive relation? – Med Oct 20 '16 at 12:36
  • @Med Apologies. It is fixed. – Taln Oct 20 '16 at 12:37
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    @Peter I noticed that, but I didn't know how to show that there is never a $x^{n-1}$ term in the polynomial other than saying it doesn't exist of how the polynomial is constructed. – Taln Oct 20 '16 at 12:39
  • It seems that $a_{n-2}=-2^{n-3}\cdot n$ (True for $2\le n\le 5000$) , but I have no proof. – Peter Oct 20 '16 at 12:50
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    @Peter The exponents of $x$ are aternatively odd and even which can be shown by using induction. Therefore depending on the parity of $n$, the polynomial is either even or odd and observing $a_{n-1} = 0$ follows. – Taln Oct 20 '16 at 12:53
  • @Taln So, you solved this part. Maybe, induction can proof my other conjecture as well. – Peter Oct 20 '16 at 12:54
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    @Peter: Your statement $a_{n-2}=-n 2^{n-3}$ follows from the explicit formula $$T_n(x) = \tfrac{n}{2} \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}(-1)^k \frac{(n-k-1)!}{k!(n-2k)!}~(2x)^{n-2k}$$ if you substitute $k=1$ $$a_{n-2}=\tfrac{n}{2}(-1)^1 \frac{(n-1-1)!}{1!(n-2)!}(2)^{n-2}=-n 2^{n-3}$$ – gammatester Oct 20 '16 at 13:43

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What is obvious is that the coefficient of $x^{n-1}$ is $0$ for $T(n)$. Because $T(n-1)$ does not have $x$ of power $n-2$, so, when multiplied by $2x$, there would be no power of $n-1$. For $T(n-2)$, we do not have a $x$ of power $n-1$, just because $T(n)$, for any $n$, is of degree $n$. So, their sum would have no $x$ of power $n-1$.

For $x^{n-2}$ a recursive formula can be found. If $\{s_i|i=\{1,2,3,...\}\}$ is the sequence of the coefficients of $x^{n-2}$ for polynomials $T(n)$, then the following relation holds.

$s_n=2s_{n-1}-2^{n-3}$

The closed form is

$-n\times 2^{n-3}$

which is achieved by iterative substitution of $s(i)$ for $i<n$.

Med
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  • How did you come up with the relation for $s_n$? – Taln Oct 20 '16 at 14:33
  • The coefficient Sn of x^(n-2), in T(n), is constructed from 2 times the coefficient S(n-1) of x^(n-3), in T(n-1), minus the coefficient of x^(n-2) in T(n-2), which is 2^(n-3). I think it would be helpful to write a few of the polynomials and then the pattern is obvious. – Med Oct 20 '16 at 15:14